Question: Solve for $z$, $- \dfrac{4z - 3}{z + 9} = \dfrac{1}{6} $
Solution: Multiply both sides of the equation by $z + 9$ $ -(4z - 3) = \dfrac{z + 9}{6} $ Multiply both sides of the equation by $6$ $ -6(4z - 3) = z + 9 $ $-24z + 18 = z + 9$ $18 = 25z + 9$ $9 = 25z$ $25z = 9$ $z = \dfrac{9}{25}$